Solve for x
x\in \left(-\infty,-\sqrt{7}-4\right)\cup \left(\sqrt{7}-4,\infty\right)
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x^{2}+8x+9=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-8±\sqrt{8^{2}-4\times 1\times 9}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 8 for b, and 9 for c in the quadratic formula.
x=\frac{-8±2\sqrt{7}}{2}
Do the calculations.
x=\sqrt{7}-4 x=-\sqrt{7}-4
Solve the equation x=\frac{-8±2\sqrt{7}}{2} when ± is plus and when ± is minus.
\left(x-\left(\sqrt{7}-4\right)\right)\left(x-\left(-\sqrt{7}-4\right)\right)>0
Rewrite the inequality by using the obtained solutions.
x-\left(\sqrt{7}-4\right)<0 x-\left(-\sqrt{7}-4\right)<0
For the product to be positive, x-\left(\sqrt{7}-4\right) and x-\left(-\sqrt{7}-4\right) have to be both negative or both positive. Consider the case when x-\left(\sqrt{7}-4\right) and x-\left(-\sqrt{7}-4\right) are both negative.
x<-\left(\sqrt{7}+4\right)
The solution satisfying both inequalities is x<-\left(\sqrt{7}+4\right).
x-\left(-\sqrt{7}-4\right)>0 x-\left(\sqrt{7}-4\right)>0
Consider the case when x-\left(\sqrt{7}-4\right) and x-\left(-\sqrt{7}-4\right) are both positive.
x>\sqrt{7}-4
The solution satisfying both inequalities is x>\sqrt{7}-4.
x<-\sqrt{7}-4\text{; }x>\sqrt{7}-4
The final solution is the union of the obtained solutions.
Examples
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4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}